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Question

In an experiment on photoelectric emission from a metallic surface, wavelength of incident light is 2×107 m and stopping potential is 2.5 V. The threshold frequency of the metal is approximately (Change of electron e=1.6×1019 C, Planck's constant h=6.6×1034 Js)

A
12×1015 Hz
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B
9×1015 Hz
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C
9×1014 Hz
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D
12×1013 Hz
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Solution

The correct option is C 9×1014 Hz
Energy of incident photon, E=hcλ

E=6.6×1034×3×1082×107=9.9×1019J=9.9×10191.6×1019eVE=6.2 eVKmax=eVs=e×2.5V=2.5 eV

According to Einstein's photoelectric equation
Kmax=hcλϕ0
Where the symbols have their usual meanings
or ϕ0=hcλkmax=6.2eV2.5eV=3.7eV
Threshold frequency,
v0=ϕ0hv0=3.7×1.6×10196.6×10340.9×1015Hz=9×1014 Hz

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